Help with basic transfinite induction proof

[psie]

TL;DR Summary

I'm working on a basic transfinite induction proof, which I'd need some help with. It concerns the construction of the Borel $\sigma$-algebra.

Some definitions:

Let $\mathcal{F}_0$ be the set of open subsets of $\mathbb R$.

Successor For any ordinal $\alpha$, let $\mathcal{F}_{\alpha+1}$ be the set of countable unions of sets $E_n\in\mathcal{F}_\alpha$ and their complements $E_m: E_m^c\in \mathcal{F}_\alpha$.

Limit For any limit ordinal $\lambda$, put $\mathcal{F}_{\alpha}=\cup_{\alpha<\lambda} \mathcal{F}_\alpha$.

Together, these form a nested sequence of sets for all ordinals, that is, $\alpha<\beta\implies \mathcal{F}_\alpha\subset\mathcal{F}_\beta$. The Borel sigma algebra is $\mathcal{F}_{\omega_1}$, where $\omega_1$ is the first uncountable ordinal.

The following statement has been left as an exercise in transfinite induction in a handout.

$\mathcal{F}_{\omega_1}\subset\mathcal{G}$ for any $\sigma$-algebra $\mathcal{G}$ containing $\mathcal{F}_0$, i.e. $\mathcal{F}_{\omega_1}$ is the minimal $\sigma$-algebra containing $\mathcal{F}_0$.

I'm looking at Wikipedia and am trying to follow their outline:

1. Show it for the base case, i.e. that $\mathcal{F}_{0}\subset\mathcal{G}$.

This is, however, trivial, since we said that $\mathcal{G}$ should contain $\mathcal{F}_{0}$.

2. Show it for successor ordinals, that is, if $\mathcal{F}_\alpha\subset\mathcal{G}$, then $\mathcal{F}_{\alpha+1}\subset\mathcal{G}$.

This follows from the fact that $\mathcal{G}$ is a $\sigma$-algebra and thus it is closed by countable unions and complements, and since $\mathcal{F}_\alpha\subset\mathcal{G}$, this gives $\mathcal{F}_{\alpha+1}\subset\mathcal{G}$. However, looking at Wikipedia, I'm unsure if this is all I have to do in the successor step. On Wikipedia, they claim that, if necessary, show also that if $\mathcal{F}_\alpha\subset\mathcal{G}$, then $\mathcal{F}_\beta\subset\mathcal{G}$ for all $\beta<\alpha$. I'm not sure if have to or how to show this.

3. Show it for limit ordinals, that is, if $\mathcal{F}_\alpha\subset\mathcal{G}$ for all $\alpha<\lambda$ where $\lambda$ is a limit ordinal, then $\mathcal{F}_\lambda\subset\mathcal{G}$.

I'm not sure how to do this.

Appreciate any help.

 

[andrewkirk]

2. ... looking at Wikipedia, I'm unsure if this is all I have to do in the successor step. On Wikipedia, they claim that, if necessary, show also that if $\mathcal{F}_\alpha\subset\mathcal{G}$, then $\mathcal{F}_\beta\subset\mathcal{G}$ for all $\beta<\alpha$. I'm not sure if have to or how to show this.

By that "if necessary" they mean that, if you can't prove
$$\mathcal{F}_{\alpha+1}\subset\mathcal{G}$$
by assuming
$$\mathcal{F}_{\alpha}\subset\mathcal{G}$$
then try proving it by assuming the stronger precedent:
$$\forall \beta \leq \alpha:\ \mathcal{F}_{\beta}\subset\mathcal{G}$$

Since you were able to do the first, you don't need to do the second. Using a weaker precedent (first case) means you have a stronger result.

 

[psie]

Thanks. I guess 3. is not so hard after all.

If $\mathcal{F}_\alpha\subset\mathcal{G}$ for all $\alpha<\lambda$, then $\mathcal{F}_\lambda=\cup_{\alpha<\lambda}\mathcal{F}_\alpha\subset\mathcal{G}$. The conclusion follows from the following implications: $$E\in \cup_{\alpha<\lambda}\mathcal{F}_\alpha\implies E\in\mathcal{F}_\alpha \text{ for some } \alpha<\lambda \implies E\in\mathcal{F}_\alpha\subset\mathcal{G} \text{ by assumption.}$$ Also the nested property $\alpha<\beta\implies \mathcal{F}_\alpha\subset\mathcal{F}_\beta$ for any ordinals $\alpha,\beta$ is not hard to show. Either $\beta$ is a successor ordinal or a limit ordinal. If it is a successor ordinal, note that $E\in\mathcal{F}_\alpha\implies E\in\mathcal{F}_{\alpha+1}$ for any $\alpha$ since every $\mathcal{F}_\alpha$ contains $\emptyset$ and $\mathbb R$ (and thus by complements and countable unions, so will $\mathcal{F}_{\alpha+1}$). If $\beta$ is a limit ordinal, then $E\in\mathcal{F}_\alpha$ implies it is in the union of all $\mathcal{F}_\alpha$ for $\alpha<\beta$, so $E\in \cup_{\alpha<\beta}\mathcal{F}_\alpha=\mathcal{F}_\beta$.